Introduction to Electrochemistry

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					  Introduction to Electrochemistry
                   Read Chapter 14.


Electrochemistry can be broadly defined as the study of
charge-transfer phenomena. As such, the field of
electrochemistry includes a wide range of different
chemical and physical phenomena. These areas include
(but are not limited to): battery chemistry,
photosynthesis, ion-selective electrodes, coulometry, and
many biochemical processes. Although wide ranging,
electrochemistry has found many practical applications
in analytical measurements.
          Electro-analytical Chemistry
Electro-analytical chemistry is the field of electrochemistry
that utilizes the relationship between chemical phenomena
which involve charge transfer (e.g. redox reactions, ion
separation, etc.) and the electrical properties that
accompany these phenomena for some analytical
determination. This relationship is further broken down into
fields based on the type of measurement that is made.
Potentiometry involves the measurement of potential for
quantitative analysis, and electrolytic electrochemical
phenomena involve the application of a potential or current
to drive a chemical phenomenon, resulting in some
measurable signal which may be used in an analytical
determination.
There are two parts to understanding electrochemistry:
the first is thermodynamics; the second is the kinetics
of electrode processes. For the latter, one needs to
study the surface chemistry to obtain a real
understanding of how electrochemical systems work.

For chemists to understand the principles underlying
the functioning of such practically important systems
as batteries, fuel cells, corrosion, electrolysis, as well
as membranes and biomembranes (of utmost
importance for the understanding of, drug delivery and
the functioning of cell membranes) they must therefore
be taught the basics of interfacial structure,
electrochemical kinetics and transport processes.
(From International Society of Electrochemistry) site
Example of a Galvanic Cell
Redox reactions involve electron transfer.
Acid-Base reactions involve proton transfer.

The key difference is that electrons can be
transported through space via wires or any other
conducting device with the result that the oxidation
and reduction reactions can occur in different places.
Protons are transported in an aqueous (or some other
polar) environment, so acid-base reactions occur in
the same place.
               Basic Concepts
Redox reactions involve a species which is
oxidized and another that is reduced.

       Fe3  V 2  Fe2  V 3

In the above, Fe3+ is reduced to Fe2+ .
It is the oxidizing agent.

Since DG<0 for this reaction we can say that V3+
wants the extra electron less than Fe3+.

               V 3  Fe3
     2           2   Galvanic Cell
Cd         2Ag
      An Aside
Why won’t this cell work?

  Ag+ will go to left
  electrode and ask for
  e from Cd(s) directly.
Will this cell work?
How badly to the electrons want to flow?




                                       q
                                           I  E/R
                                      sec

                                I = current in amps
                                R = resistance in ohms
                                E = potential difference in
        q = n      x F              Volts

Coulombs  Moles 
                   Coulomb
                                    DG  -nFE
                    Moles
                      Voltaic Cells
Electrochemical cells that use an oxidation-reduction
reaction to generate an electric current are known as
galvanic or voltaic cells.


Let's take another look at the voltaic cell in the figure
below.
             Voltaic Cells




Taken from http://chemed.chem.purdue.edu/genchem
Within each half-cell, reaction occurs on the surface of the
metal electrode. At the zinc electrode, zinc atoms are
oxidized to form Zn2+ ions, which go into solution. The
electrons liberated in this reaction flow through the zinc
metal until they reach the wire that connects the zinc
electrode to the platinum wire. They then flow through the
platinum wire, where they eventually reduce an H+ ion in
the neighboring solution to a hydrogen atom, which
combines with another hydrogen atom to form an H2
molecule.
The electrode at which oxidation takes place in a
electrochemical cell is called the anode. The electrode at
which reduction occurs is called the cathode. The
identity of the cathode and anode can be remembered by
recognizing that positive ions, or cations, flow toward
the cathode, while negative ions, or anions, flow toward
the anode. In the voltaic cell shown above, H+ ions flow
toward the cathode, where they are reduced to H2 gas. On
the other side of the cell, Cl- ions are released from the
salt bridge and flow toward the anode, where the zinc
metal is oxidized.
     The voltaic cell consist of the two reactions.
         Zn(s)  Zn 2  2e             oxidation
   + 2H   2e -  H (g)                 reduction
                      2

      2H   Zn(s)  Zn 2  H 2 (g)
 Or equivalently we can write the reactions as follows
           2H   2e -  H 2 (g)
          Zn 2  2e   Zn(s)

      2H   Zn(s)  Zn 2  H 2 (g)
We can only measure E for the full reaction.
We would like to calculate E for the half reactions.
Before doing this, we must recognize the E depends on
concentrations.
               Voltaic Cells



                                                    Zn 2   1
                                                   H  1
                                                    H2 (g)  1




Taken from http://chemed.chem.purdue.edu/genchem
Since reactants and products are in their standard states, we
call the E for this cell the standard reduction potential (Eo).
Here Eo = .76V.
                       2          
         Zn(s)  Zn          2e
        +   2H   2e -  H 2 (g)

        2H   Zn(s)  Zn 2  H 2 (g)
We arbitrarily define the potential for, one half reaction, the
second reaction above to be exactly 0V when reactants and
products are in their standard states.

Since Eo for the cell is the sum of Eo’s for the two half
reactions we see that Eo for the first half reaction is .76V.
Oxidizing Power Increases
               Voltaic Cells



                                                    Zn 2   1
                                                   H  1
                                                    H2 (g)  1




Taken from http://chemed.chem.purdue.edu/genchem
                    Zn(s)  Zn 2  2e 
                   +   2H   2e -  H 2 (g)
                                        2
                  2H  Zn(s)  Zn              H 2 (g)


This voltaic cell on the previous slide is fully described with
the following notation

Zn(s) | Zn 2 (aq, Α  1) || H  (aq, Α  1) | H 2 (g, Α  1) | Pt(s)

Zn(s) | Zn 2 (aq, Α  1) || S.H.E.
           Line Notation For Voltaic Cells
Voltaic cells can be described by a line notation based on the
following conventions.
    Single vertical line indicates change in state or phase.
    Within a half-cell, the reactants are listed before the
        products.
    Activities of aqueous sol’ns are written in parentheses
    after the symbol for the ion or molecule.
    A double vertical line indicates a junction between half-
    cells.
    The line notation for the anode (oxidation) is written
    before the line notation for the cathode (reduction).
The line notation for a standard-state Daniell cell is written
as follows.
Electrons flow from the anode to the cathode in a voltaic
cell. (They flow from the electrode at which they are given
off to the electrode at which they are consumed.) Reading
from left to right, this line notation therefore corresponds to
the direction in which electrons flow.




Zn    | Zn2+(1.0 M)          || Cu2+(1.0 M)          | Cu
            anode                   cathode
          (oxidation)             (reduction)
       The Nernst Equation (14-4 of book)
The Nernst equation relates the potential of a cell in its
standard state to that of a cell not in its standard state.
Consider the reaction below.
  Zn 2  2e   Zn(s)             2     
                              2Zn  4e  2Zn(s)
We know from Le Chatelier’s principle that increasing
the concentration of Zn2+ should drive the reaction to the
right. In other words it should decrease the potential of
the half cell. The Nernst equation allows us to calculate
this increase for the above half reaction as
      RT            Zn(s)            RT 
                                                     2

EE   o
         ln                     EE   o
                                        ln
                                                     Zn(s)

      2F  Zn 2                      4F            2
                                                     Zn 2
       The Nernst Equation Continued
The Nernst Eq. for the reaction bB  ne   cC   is

                      RT         c
                EE  o
                        ln        C

                      nF         b
                                  B

                             At 25oC this equation
                             simplifies to

                  .05916V            c
            EE  o
                          log         C

                      n              b
                                      B
 The Nernst Equation For Complete Cell
                    E  E  E
Here E+ and E- are the potentials of the half cells
connected to the positive and negative terminals of
potentiometer respectively. Let’s consider an
example.
               Voltaic Cells
           -                +
                                                    Zn 2   1
                                                   H  1
                                                    H2 (g)  1




Taken from http://chemed.chem.purdue.edu/genchem
                   E  E  E
 E+ and E- are potentials of half cells connected to positive
 and negative terminals of potentiometer respectively
                  2H   2e -  H 2 (g)
                  Zn 2  2e   Zn(s)

              2H   Zn(s)  Zn 2  H 2 (g)

     .05916V    H2 ( g )      .05916V      Zn(s) 
E          log 2  -  - .76          log         
         2       H               2        Zn 2 
                                                   
                 .05916      H 2 ( g )  Zn 2   See page 295 for
       E  .76         log                       another example
                    2            2  H
       Calculating Equilibrium Constants
Consider the reaction Ce 4  Fe2  Ce3  Fe3
made up of the following two half reactions

 Ce 4  e  Ce3             Eo=1.700V
     3        2
  Fe  e  Fe                  Eo=0.767V

 Since Eo is greater for cerium this reaction will be the reduction
 reaction. The standard potential for the galvanic cell would be

             E  E - E  1.700 - (0.767)
               o      o
                      
                           o
                           
       Calculating Equilibrium Constants
                   Continued
Consider the reaction Ce 4  Fe2  Ce3  Fe3
In a galvanic cell we would have
                                    3     3
                           RT [Fe ][Ce ]
     E  E - E-  E - E -
                   0
                   
                        0
                              ln
                            nF [Fe 2 ][Ce 4 ]
 At equilibrium E=0 and
          RT [Fe 3 ][Ce 3 ]
Eo - Eo 
            ln    2    4
                               .05916  log K at 25 o C
           F     [Fe ][Ce ]
                                      K  1016
                 DG o
     Eo - Eo 
          
                              This connection to free
                  nF          energy is important
    Calculating Equilibrium Constants for
         Nonredox Reactions (14-5)
                                       2
Consider the reaction FeCO 3 ( s )  Fe  CO 3
                                             2-


This is a Ksp problem. Not a redox problem. Nonetheless
we can use electrochemistry to calculate Ksp by considering
                   
    FeCO3 ( s)  2e  Fe(s)  CO       2-
                                       3    E  0.756V
                                             o
                                             

               Fe2  2e   Fe(s)          E o  0.44V
                                              


  E -E 
    o
    
        o
        
            .059
              2
                       2
                 log [Fe ][CO3 ] 
                              2-
                                   .059
                                      2
                                             log K sp
  2- .316                      2.316 / .059      11
             log K sp  Ksp  10                  10
    .059
                                                 (at 25oC)
             Latimer Diagrams (Box 14-2 of book)
 The oxidation states of elements are related to each other

IO   IO   HOI   I 2  I
  -
  4        
            1.589
                              -
                               3
                                       1.154              1.430          .535      -

(+7)                        (+5) 1.299         (+1)               (0)            (-1)
                    
            2H  IO  2e  IO  H 2 O
                               -
                               4
                                                     -
                                                      3
                                                                        DG   o
                                                                             1

           5H   IO 3  4e   HOI  2H 2 O
                      -
                                                                        DG o
                                                                           2

           7H   IO -4  6e   HOI  3H 2O                            DG   o
                                                                             3

                         2 1.589  4 1.154
DG  DG  DG
       o
       1
                        o
                        2
                                   o
                                   3
                                              1.299
                                  6
                               2E 1  4E o
                                  o
n1FE1  n 2 FE 2  n 3FE 3
    o          o         o                2
                                             E3
                                               o

                                    6
              Electrochemistry Skills
• Understand how voltaic cells work.
• Be able to calculate standard reduction potentials for
  voltaic cells, given the chemical reactions.
• Be able to describe a voltaic cell using the line notation
  and visa versa. Know which way electrons flow and
  where the anode and cathode are.
• Know how to work with the Nernst Eq. to include
  concentration dependencies and calculate equilibirum
  constants.
• Know the relation between E and DG and can use this
  relation in constructing Latimer diagrams.

				
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